Multiply the following complex numbers: $({-3-i}) \cdot ({3+i})$
Solution: Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({-3-i}) \cdot ({3+i}) = $ $ ({-3} \cdot {3}) + ({-3} \cdot {1}i) + ({-1}i \cdot {3}) + ({-1}i \cdot {1}i) $ Then simplify the terms: $ (-9) + (-3i) + (-3i) + (-1 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ -9 + (-3 - 3)i - 1i^2 $ After we plug in $i^2 = -1$ , the result becomes $ -9 + (-3 - 3)i - (-1) $ The result is simplified: $ (-9 + 1) + (-6i) = -8-6i $